Factorisation

∴ Given terms are 8x and 24

Prime factors of given terms are:-

8x = 2 × 2 × 2 × x

24 = 2 × 2 × 2 × 3

As the x is an undefined value,

Common factors will be

2 × 2 × 2 = 8

∴ Given terms are 3a and 21ab

Prime factors of given terms are:-

3a = 3 × a

21ab = a × b × 7 × 3

Common factors will be

⇒ 3 × a = 3a

∴ Given terms are 7xy and 35x²y³

Prime factors of given terms are:-

7xy = 7 × x × y

35x²y³ = x × x × y × y × 7 × 5

Common factors will be

⇒ 7 × x × y = 7xy

∴ Given terms are 4m²,6m² and 8m³

Prime factors of given terms are:-

4m² = 2 × 2 × m × m

6m² = 3 × 2 × m × m

8m³ = 2 × 2 × 2 × m × m × m

Common factors will be

⇒ 2 × m × m = 2m²

∴ Given terms are 15p,20qr and 25rp

Prime factors of given terms are:-

15p = 3 × 5 × p

20qr = 2 × 2 × 5 × q × r

25rp = 5 × 5 × r × p

⇒ Common factors will be 5

∴ Given terms are 4x²,6xy and 8y²x

Prime factors of given terms are:-

4x² = 2 × 2 × x × x

6xy = 3 × 2 × x × y

8y²x = 2 × 2 × 2 × y × y × x

Common factors will be

⇒ 2 × x = 2x

Given terms are 12x²yand 18xy²

Prime factors of given terms are:-

12yx² = 3 × 2 × 2 × y × x × x

18xy² = 3 × 2 × 3 × x × y × y

Common factors will be

⇒ 2 × 3 × x × y = 6xy

In the given expression

Check the common factors for all terms;

⇒ [5 × x × x - 5 × 5 × x × y]

⇒ 5 × x[x-5 × y]

⇒ 5x[x-5y]

∴ 5x² - 25xy = 5x[x-5y]

In the given expression

Check the common factors for all terms;

⇒ [5 × a × a- 2 × 3 × x × a]

⇒ a[5 × a-2 × 3 × x]

⇒ a[5a-6x]

∴ 9a² - 6ax = a[5a-6x]

In the given expression

Check the common factors for all terms;

⇒ [7 × p × p + 7 × 7 × p × q]

⇒ 7 × p[p + 7 × q]

⇒ 7p[p + 7q]

∴ 7p² + 49pq = 7p[p + 7q]

In the given expression

Check the common factors for all terms;

⇒ [2 × 2 × 3 × 3 × a × a × b - 2 × 2 × 3 × 5 × a × a × b × c]

⇒ 2 × 2 × 3 × a × a × b[3 × b-5 × c]

⇒ 12a²b[3b-5c]

∴ 36a²b - 60 a²bc = 12a²b[3b-5c]

In the given expression

Check the common factors for all terms;

⇒ [3 × a × a × b × c + 2 × 3 × a × b × b × c + 3 × 3 × a × b × c × c]

⇒ 3 × a × b × c[a + 2 × b + 3 × c]

⇒ 3abc[a + 2b + 3c]

∴ 3a²bc + 6ab²c + 9abc² = 3abc[a + 2b + 3c]

In the given expression

Check the common factors for all terms;

⇒ [2 × 2 × p × p + 5 × p × q - 2 × 3 × p × q × q]

⇒ p[2 × 2 × p + 5 × q - 2 × 3 × q × q]

⇒ p[4p + 5q-6q²]

∴ 4p² + 5pq – 6pq² = p[4p + 5q-6q²]

In the given expression

Check the common factors for all terms;

⇒ [u × t + a × t × t]

⇒ t[u + a × t]

⇒ t[u + at]

∴ ut + at² = t[u + at]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

3ax-6xy = 3x[a-2y] -------eq 1

Regrouping the last 2 terms we have,

8by-4ab = -4b[a-2y] -------eq 2

∵ we have to make common parts in both eq 1 and 2

Combining eq 1 and 2

3ax – 6xy + 8by – 4ab = 3x[a-2y] + [-4b[a-2y] ]

= 3x[a-2y] - 4b[a-2y]

= [3x-4] [a-2y]

Hence the factors of 3ax – 6xy + 8by – 4ab are [3x-4] and [a-2y]

In the given expression

Check whether there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

x³ + 2x² = x²[x + 2] -------eq 1

Regrouping the last 2 terms we have,

5x + 10 = 5[x + 2] -------eq 2

Combining eq 1 and 2

x³ + 2x² + 5x + 10 = x²[x + 2] + 5[x + 2]

= [x² + 5][x + 2]

Hence the factors of x³ + 2x² + 5x + 10 are [x² + 5] and [x + 2]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

m² - mn= m[m - n] -------eq 1

Regrouping the last 2 terms we have,

4m – 4n = 4[m – n] -------eq 2

Combining eq 1 and 2

m² – mn + 4m – 4n = 4[m – n] + m[m - n]

= [4 + m][m-n]

Hence the factors of m² – mn + 4m – 4n are [m – n] and [4 + m]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

a³ – a²b² = a²[a-b²] -------eq 1

Regrouping the last 2 terms we have,

– ab + b³ = -b[a-b²] -------eq 2

∵ we have to make common parts in both eq 1 and 2

Combining eq 1 and 2

a³ – a²b² – ab + b³ = a²[a-b²] -b[a-b²]

= [a² – b][a – b²]

Hence the factors of a³ – a²b² – ab + b³ are [a² – b] and[a – b²]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

p²q – pr² = p[pq-r²] -------eq 1

Regrouping the last 2 terms we have,

– pq + r² = -1[pq-r²] -------eq 2

∵ we have to make common parts in both eq 1 and 2

Combining eq 1 and 2

p²q – pr² – pq + r² = p[pq-r²] -1[pq-r²]

= [p – 1][pq – r²]

Hence the factors of p²q – pr² – pq + r² are [p – 1] and [pq – r²]

In the given expression

1^st and last terms are perfect square

⇒ a² = a × a

⇒ 25 = 5 × 5

And the middle expression is in form of 2ab

10a = 2 × 5 × a

∴ a × a + 2 × 5 × a + 5 × 5

Gives (a + b)² = a² + 2ab + b²

⇒ In a² + 10a + 25

a = a and b = 5;

∴ a² + 10a + 25 = (a + 5)²

Hence the factors of a² + 10a + 25 are (a + 5) and (a + 5)

In the given expression

1^st and last terms are perfect square

⇒ l² = l × l

⇒ 64 = 8 × 8

And the middle expression is in form of 2ab

16l = 2 × 8 × l

∴ l × l + 2 × 8 × l + 8 × 8

Gives (a-b)² = a²-2ab + b²

⇒ In l² + 16l + 64

a = l and b = 8;

∴ l² + 16l + 64 = (l + 8)²

Hence the factors of l² + 16l + 64 are (l + 8) and (l + 8)

In the given expression

1^st and last terms are perfect square

⇒ 36x² = 6x × 6x

⇒ 64y² = 8y × 8y

And the middle expression is in form of 2ab

96xy = 2 × 6x × 8y

∴ 6x × 6x + 2 × 8y × 6x + 8y × 8y

Gives (a + b)² = a² + 2ab + b²

⇒ In 36x² + 96xy + 64y²

a = 6x and b = 8y;

∴ 36x² + 96xy + 64y² = (6x + 8y)²

Hence the factors of 36x² + 96xy + 64y² are (6x + 8y) and (6x + 8y)

In the given expression

1^st and last terms are perfect square

⇒ 25x² = 5x × 5x

⇒ 9y² = 3y × 3y

And the middle expression is in form of 2ab

30xy = 2 × 5x × 3y

∴ 5x × 5x + 2 × 3y × 5x + 3y × 3y

Gives (a-b)² = a²-2ab + b²

⇒ In 25x² – 30xy + 9y²

a = 5x and b = 3y;

∴ 25x² - 30xy + 9y² = (5x-3y)²

Hence the factors of 25x² - 30xy + 9y² are (5x-3y) and (5x-3y)

In the given expression

1^st and last terms are perfect square

⇒ 25m² = 5m × 5m

⇒ 16n² = 4n × 4n

And the middle expression is in form of 2ab

40mn = 2 × 5m × 4n

∴ 5m × 5m - 2 × 4n × 5m + 4n × 4n

Gives (a-b)² = a²-2ab + b²

⇒ In 25m² – 40mn + 16n²

a = 5m and b = 4n;

∴ 25m² – 40mn + 16n² = (5m-4n)²

Hence the factors of 25m² – 40mn + 16n² are (5m-4n)and (5m-4n)

In the given expression

1^st and last terms are perfect square

⇒ 81x² = 9x × 9x

⇒ 121y² = 11y × 11y

And the middle expression is in form of 2ab

198xy = 2 × 9x × 11y

∴ 9x × 9x - 2 × 11y × 9x + 11y × 11y

Gives (a-b)² = a²-2ab + b²

⇒ In 81x² – 198xy + 121y²

a = 9x and b = 11y;

∴ 81x² – 198xy + 121y² = (9x-11y)²

Hence the factors of 81x² – 198xy + 121y² are (9x-11y)and (9x-11y)

If (a + b)² = a² + 2ab + b²

Then (x + y)² – 4xy

= x² + 2xy + y²-4xy

= x² + y²-2xy

In given expression

1^st and last terms are perfect square

⇒ x² = x × x

⇒ y² = y × y

And the middle expression is in form of 2ab

2xy = 2 × x × y

∴ x × x - 2 × y × x + y × y

Gives (a-b)² = a²-2ab + b²

⇒ In x² – 2xy + y²

a = x and b = y;

∴ x² – 2xy + y² = (x-y)²

Hence the factors of (x + y)² – 4xy are (x-y)and (x-y)

In given expression

1^st and last terms are perfect square

⇒ l⁴ = l² × l²

⇒ m⁴ = m² × m²

And the middle expression is in form of 2ab

4l²m² = 2 × l² × m²

∴ l² × l² + 2 × m² × l² + m² × m²

Gives (a + b)² = a² + 2ab + b²

⇒ In l⁴ + 4l²m² + m⁴

a = l² and b = m²;

∴ l⁴ – 4l²m² + m⁴ = (l²-m²)²

Hence the factors of l⁴ + 4l²m² + 4m⁴ are (l²-m²)and (l²-m²)

In given expression

Both terms are perfect square

⇒ x² = x × x

⇒ 36 = 6 × 6

∴ x²-6²

Seems to be in identity a²-b² = (a + b)(a-b)

Where a = x and b = 6;

x² – 36 = (x + 6)(x-6)

Hence the factors of x² – 36 are (x + 6) and (x-6)

In given expression

Both terms are perfect square

⇒ 49x² = 7x × 7x

⇒ 25y² = 5y × 5y

∴ 49x²-25y² Seems to be in identity a²-b² = (a + b)(a-b)

Where a = 7x and b = 5y;

49x² – 25y² = (7x + 5y)(7x-5y)

Hence the factors of 49x² – 25y² are (7x + 5y) and (7x-5y)

In given expression

Both terms are perfect square

⇒ m² = m × m

⇒ 121 = 11 × 11

∴ m²-121 Seems to be in identity a²-b² = (a + b)(a-b)

Where a = m and b = 11;

m² – 121 = (m + 11)(m-11)

Hence the factors of m² – 121 are (m + 11) and (m-11)

In given expression

Both terms are perfect square

⇒ 64x² = 8x × 8x

⇒ 81 = 9 × 9

∴ 81-64x² Seems to be in identity a²-b² = (a + b)(a-b)

Where a = 9 and b = 8x;

81-64x² = (9-8x)(9 + 8x)

Hence the factors of 81-64x²are(9-8x) and (9 + 8x)

In given expression

Both terms are perfect square

⇒ y²x² = xy × xy

⇒ 64 = 8 × 8

∴ x²y² – 64 Seems to be in identity a²-b² = (a + b)(a-b)

Where a = xy and b = 8;

x²y² – 64 = (xy-8)(xy + 8)

Hence the factors of x²y² – 64 are (xy-8) and (xy + 8)

In given expression

Take out the common factor,

[2 × 3 × x × x-2 × 3 × 3 × 3]

⇒ 2 × 3[x × x-3 × 3]

⇒ 6[x²-9]

Both terms are perfect square

⇒ x² = x × x

⇒ 9 = 3 × 3

∴ x²– 9 Seems to be in identity a²-b² = (a + b)(a-b)

Where a = x and b = 3;

x²– 9 = (x-3)(x + 3)

Hence the factors of 6x² – 54 are 6,(x-3) and (x + 3)

In given expression

Both terms are perfect square

⇒ x² = x × x

⇒ 81 = 9 × 9

∴ x² – 81 Seems to be in identity a²-b² = (a + b)(a-b)

Where a = x and b = 9;

x² – 81 = (x-9)(x + 9)

Hence the factors of x² – 81 are (x-9) and (x-9)

In given expression

Take out the common factor,

[2 × x - 2 × 2 × 2 × 2 × 2 × x × x × x × x × x]

⇒ 2 × x[1 - 2 × 2 × 2 × 2 × x × x × x × x]

⇒ 2x [1-16x⁴] = 2x [1-(2x)⁴]

⇒ In the term 1-(2x)⁴

= 1-(4x²)²

Both terms are perfect square

⇒ (4x² )² = 4x² × 4x²

⇒ 1 = 1 × 1

∴ 1-(4x²)² Seems to be in identity a²-b² = (a + b)(a-b)

Where a = 1 and b = 4x²;

1-16x⁴ = (1-4x²)(1 + 4x²)

→ 1-4x² = 1-(2x)²

∴ 1-4x² Seems to be in identity a²-b² = (a + b)(a-b)

Where a = 1 and b = 2x;

1-4x² = (1-2x)(1 + 2x)

∴ 1-16x² = (1-2x)(1 + 2x) (1 + 4x²)

Hence the factors of 2x – 32x⁵ are 2x,(1-2x),(1 + 2x) and (1 + 4x²)

In given expression

Take out the common factor,

[3 × 3 × 3 × 3 × x × x × x × x - 11 × 11 × x × x]

⇒ x × x[3 × 3 × 3 × 3 × x × x - 11 × 11]

⇒ x²[81x² – 121]

In expression 81x² - 121

Both terms are perfect square

⇒ 81x² = 9x × 9x

⇒ 121 = 11 × 11

∴ 81x² – 121 Seems to be in identity a²-b² = (a + b)(a-b)

Where a = 9x and b = 11;

81x² – 121 = (9x-11)(9x + 11)

Hence the factors of 81x⁴ – 121x² are x²,(9x-11) and (9x + 11)

In the given expression p² – 2pq + q²

1^st and last terms are perfect square

⇒ p² = p × p

⇒ q² = q × q

And the middle expression is in form of 2ab

2pq = 2 × p × q

∴ p × p - 2 × p × q + q × q

Gives (a-b)² = a²-2ab + b²

⇒ In p² – 2pq + q²

a = p and b = q;

∴ p² – 2pq + q² = (p-q)²

Now the given expression is (p-q)²– r²

Both terms are perfect square

⇒ (p-q)² = (p-q) × (p-q)

⇒ r² = r × r

∴ (p-q)²– r² Seems to be in identity a²-b² = (a + b)(a-b)

Where a = (p-q) and b = r;

(p-q)²– r² = (p-q-r) (p-q + r)

Hence the factors of (p² – 2pq + q²) – r² are (p-q-r) and (p-q + r)

In the given expression

We know that

(a + b)² = a² + 2ab + b²

(a-b)² = a²-2ab + b²

Hence

If a = x and b = y

(x + y)² – (x – y)² = x² + y² + 2xy – [x² + y²-2xy]

= x² + y² + 2xy -x²-y² + 2xy

= 4xy

In the given expression

Take out the common in all the terms,

⇒ lx² + mx

⇒ x[lx + m]

In the given expression

Take out the common in all the terms,

⇒ 7y² + 35z²

⇒ 7[y² + 5z²]

In the given expression

Take out the common in all the terms,

⇒ 3x⁴ + 6x³y + 9x²z

⇒ 3x²[x² + 2xy + 3z]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

x² - ax= x[x - a] -------eq 1

Regrouping the last 2 terms we have,

-bx + ab = -b[x – a] -------eq 2

Combining eq 1 and 2

x² – ax – bx + ab = x[x - a] - b[x – a]

= [x - b][x - a]

Hence the factors of[ x² – ax – bx + ab] are [x - b]and [x - a]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

3ax – 6ay= 3a[x - 2y] -------eq 1

Regrouping the last 2 terms we have,

-8by + 4bx = 4b[x – 2y] -------eq 2

Combining eq 1 and 2

3ax – 6ay – 8by + 4bx = 3a[x - 2y] + 4b[x – 2y]

= [x – 2y][3a + 4b]

Hence the factors of[3ax – 6ay – 8by + 4bx] are [x – 2y] and [3a + 4b]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

mn + m = m[n + 1] -------eq 1

Regrouping the last 2 terms we have,

n + 1 = 1[n + 1] -------eq 2

Combining eq 1 and 2

mn + m + n + 1 = m[n + 1] + 1[n + 1]

= [m + 1][n + 1]

Hence the factors of[mn + m + n + 1] are [m + 1] and [n + 1]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

6ab – b² = b[6a - b] -------eq 1

Regrouping the last 2 terms we have,

12ac – 2bc = 2c[6a - b] -------eq 2

Combining eq 1 and 2

6ab – b² + 12ac – 2bc = b[6a - b] + 2c[6a - b]

= [6a - b][b + 2c]

Hence the factors of[6ab – b² + 12ac – 2bc] are [6a - b] and[b + 2c]

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1^st 2 terms we have,

p²q – pr² = p[pq – r²] -------eq 1

Regrouping the last 2 terms we have,

– pq + r² = -1[pq – r²] -------eq 2

∵ we have to make common parts in both eq 1 and 2

Combining eq 1 and 2

p²q – pr² – pq + r² = p[pq – r²] -1[pq – r²]

= [pq – r²][p - 1]

Hence the factors of[p²q – pr² – pq + r²] are [pq – r²] and [p - 1]

In the given expression

Take out the common in all the terms,

⇒ x (y + z) – 5 (y + z)

⇒ (y + z)(x - 5)

Hence the factors of x (y + z) – 5 (y + z) are (y + z) and (x - 5)

In expression x⁴ – y⁴

Both terms are perfect square

⇒ x⁴ = x² × x²

⇒ y⁴ = y² × y²

∴ x⁴ – y⁴ Seems to be in identity a²-b² = (a + b)(a-b)

Where a = x² and b = y²;

x⁴ – y⁴ = (x² – y²)( x² + y²),

∴ x² – y² Seems to be in identity a²-b² = (a + b)(a-b)

Where a = x and b = y;

x² – y² = (x– y)( x+ y),

⇒ x⁴ – y⁴ = (x– y)( x+ y), ( x² + y²)

Hence the factors of x⁴ – y⁴ are (x– y),( x+ y) and ( x² + y²)

In expression a⁴ – (b + c)⁴

Both terms are perfect square

⇒ a⁴ = a² × a²

⇒ (b + c)⁴ = (b + c)² × (b + c)²

∴ a⁴ – (b + c)⁴ Seems to be in identity a²-b² = (a + b)(a-b)

Where a = a² and b = (b + c)²;

a⁴ – (b + c)⁴ = (a² – (b + c)²)( a² + (b + c)²),

∴ a² – (b + c)² Seems to be in identity a²-b² = (a + b)(a-b)

Where a = a and b = (b + c);

a² – (b + c)² = (a– (b + c))( a+ (b + c)),

⇒ a⁴ – (b + c)⁴ = (a– (b + c))( a+ (b + c)), ( a² + (b + c)²)

⇒ a⁴ – (b + c)⁴ = (a–b–c)(a + b + c), (a² + b² + c² + 2bc)

Hence the factors of a⁴ – (b + c)⁴ are (a–b–c),(a + b + c),( a² + b² + c² + 2bc)

In the given expression l² – (m – n)²

Both terms are perfect square

⇒ l² = l × l

⇒ (m – n)² = (m – n) × (m – n)

∴ l² – (m - n)² Seems to be in identity a²-b² = (a + b)(a-b)

Where a = a and b = (m - n);

∴ l² – (m – n)² = (l + m-n)(l-m + n)

Hence the factors of l²–(m–n)²are (l + m-n)(l-m + n)

In the given expression 49x² –

Both terms are perfect square

⇒ 49x² = 7x × 7x

⇒ ()² =

49x² – Seems to be in identity a²-b² = (a + b)(a-b)

Where a = 7x and b = ;

∴ (7x)² –()² = ( 7x – ) (7x + )

Hence the factors of 49x² – are ( 7x – ) and (7x + )

In the given expression

1^st and last terms are perfect square

⇒ x⁴ = x² × x²

⇒ y⁴ = y² × y²

And the middle expression is in form of 2ab

2x²y² = 2 × x² × y²

∴ x² × x² - 2 × x² × y² + y² × y²

Gives (a-b)² = a²-2ab + b²

⇒ x⁴ – 2x²y² + y⁴

a = x² and b = y²;

∴ x⁴ – 2x²y² + y⁴ = (x² – y²) (x² + y²)

Hence the factors of x⁴ – 2x²y² + y⁴ are (x² – y²) and (x² + y²)

In the given expression

We know that

(a + b)² = a² + 2ab + b²

(a-b)² = a²-2ab + b²

Hence

4[a² + 2ab + b²] – 9[a²-2ab + b²]

4a² + 8ab + 4b² - 9a² + 18ab - 9b²

26ab – 5a² - 5b²

25ab + ab – 5a² – 5b²

[25ab – 5a²] + [ab – 5b²]

5a[5b – a] – b[5b – a]

[5a – b][5b – a]

Hence the factors 4 (a + b)² – 9 (a – b)² are [5a – b] and [5b – a]

The given expression looks as

x² + (a + b)x + ab

where a + b = 10; and ab = 24;

factors of 24 their sum

1 × 24 1 + 24 = 25

12 × 2 2 + 12 = 14

6 × 4 6 + 4 = 10

∴ the factors having sum 10 are 6 and 4

a² + 10a + 24 = a² + (6 + 4)a + 24

= a² + 6a + 4a + 24

= a(a + 6) + 4(a + 6)

= (a + 6)(a + 4)

Hence the factors of a² + 10a + 24 are (a + 6) and (a + 4)

The given expression looks as

x² + (a + b)x + ab

where a + b = 9; and ab = 18;

factors of 18 their sum

1 × 18 1 + 18 = 19

9 × 2 2 + 9 = 11

6 × 3 6 + 3 = 9

∴ the factors having sum 9 are 6 and 3

x² + 9x + 18 = x² + (6 + 3)x + 18

= x² + 6x + 3x + 18

= x(x + 6) + 3(x + 6)

= (x + 6)(x + 3)

Hence the factors of x² + 9x + 18 are (x + 6) and (x + 3)

The given expression looks as

x² + (a + b)x + ab

where a + b = -10; and ab = 21;

factors of 21 their sum

-1 × -21 -1-18 = -19

-7 × -3 -7-3 = -10

∴ the factors having sum -10 are -7 and -3

p² + 9p + 18 = p² + (-7-3)p + 21

= p² -7p-3p + 21

= p(p-7) -3(p-7)

= (p-7)(p-3)

Hence the factors of p² + 9p + 18 are (p-7) and (p-3)

The given expression looks as

x² + (a + b)x + ab

where a + b = -4; and ab = -32;

factors of -32 their sum

1 × -32 1-32 = -31

-16 × 2 2 -16 = - 14

-8 × 4 4 -8 = -4

∴ the factors having sum -4 are -8 and 4

x² – 4x – 32 = x² + (4 -8)x - 32

= x² + 4x - 8x - 32

= x(x + 4) -8(x + 4)

= (x + 4)(x-8)

Hence the factors of x² – 4x – 32 are (x + 4) and (x-8)

A = √ s(s-a)(s-b)(s-c)

If the area is an integer

Then [s(s-a)(s-b)(s-c)] should be proper square

If s = Then s = = 24

Hence ;

A = √ 24(24-a)(24-b)(24-c)

If side of triangle are

a = 21 and b + c = 27

let c be smallest side

then b = 27-c

∴ √ 24(24-21)(24-27 + c)(24-c)

⇒ √ 24 × 3 × (c-3)(24-c)

⇒ √ 2 × 2 × 2 × 3 × 3 × (c-3)(24-c)

⇒ 2 × 3√2(c-3)(24-c)

⇒ 6√2(c-3)(24-c)

∴ the value of [2(c-3)(24-c)] must be a perfect square for area to be a integer

For getting square 2(c-3) should be equal to (24-c)

2(c-3) = (24-c)

2c-6 = 24-c

2c + c = 24 + 6

3c = 10

c = 10; b = 27-c = 27-10 = 17

Hence the size of smallest size is 10.

For the given 2 degree equation

That must be equal to(ax + by + c)(dx + e)

= ad.x² + bd.xy + cd.x + ea.x + be.y + ec

= ad.x² + bd.xy + (cd + ea).x + be.y + ec

x² + 3xy + x + my–m = ad.x² + bd.xy + (cd + ea).x + be.y + ec

compare the equation

and take out the coefficient of every term

a.d = 1 ----------1

b.d = 3 ----------2

c.d + e.a = 1 ----------3

b.e = m ----------4

e.c = -m ----------5

⇒ from eq 1; a = d = 1 ∵ all coefficient are integers

After putting result in eq 3; c + e = 1 -------6

After putting result in eq 2; b = 3 --------7

⇒ divide eq 4 and 5

∴ that implies b = -c = -3 ∵ eq 7

Put value of c in eq 6

-3 + e = 1

e = 1 + 3 = 4

Putting value of b and e in eq 4

m = b × e

m = 3 × 4 = 12

In the given term

Dividend = 48a³ = 2 × 2 × 2 × 2 × 3 × a × a × a

Divisor = 6a = 2 × 3 × a

= 2 × 2 × 2 × a × a

= 8a²

Hence dividing 48a³ by 6a gives 8a²

In the given term

Dividend = 14x³ = 2 × 7 × x × x × x

Divisor = 42x² = 2 × 3 × 7 × x × x

=

=

Hence dividing 14x³ by 42x² gives

In the given term

Dividend = 72a³b⁴c⁵ = 2 × 2 × 2 × 3 × 3 × a × a × a × b × b × b × b × c × c × c × c × c

Divisor = 8ab²c³ = 2 × 2 × 2 × a × b × b × c × c × c

=

= 3 × 3 × a × a × b × b × c × c

= 9a²b²c²

Hence dividing 72a³b⁴c⁵ by 8ab²c³ gives 9a²b²c²

In the given term

Dividend = 11xy²z³ = 11 × x × y × y × z × z × z

Divisor = 55xyz = 5 × 11 × x × y × z

=

=

=

Hence dividing 11xy²z³ by 55xyz gives

In the given term

Dividend = -54l⁴m³n² = (-1) × 2 × 3 × 3 × 3 × l × l × l × l × m × m × m × n × n

Divisor = 9l²m²n² = 3 × 3 × l × l × m × m × n × n

=

= (-1) × 3 × 2 × l × l × m

= -6l²m

Hence dividing –54l⁴m³n² by 9l²m²n² gives -6l²m

In the given term

Dividend = (3x²– 2x)

Take out the common part in binomial term

= (3 × x × x– 2 × x)

= x(3x-2)

Divisor = x

=

= 3x-2

Hence dividing (3x²– 2x) by x gives out 3x-2

In the given term

Dividend = (5a³b – 7ab³)

Take out the common part in binomial term

= (5 × a × a × a × b – 7 × a × b × b × b)

= ab(5a² – 7b²)

Divisor = ab

=

= (5a² – 7b²)

Hence dividing (5a³b – 7ab³) by ab gives out (5a² – 7b²)

In the given term

Dividend = (25x⁵ – 15x⁴)

Take out the common part in binomial term

= (5 × 5 × x × x × x × x × x – 3 × 5 × x × x × x × x)

= (5x – 3)5x⁴

Divisor = 5x³

=

= (5x – 3)x

= 5x² – 3x

Hence dividing (25x⁵ – 15x⁴) by 5x³ gives out 5x² – 3x

In the given term

Dividend = (4l⁵ – 6l⁴ + 8l³)

Take out the common part in binomial term

= (2 × 2 × l × l × l × l × l– 3 × 2 × l × l × l × l + 2 × 2 × 2 × l × l × l )

= (2l² – 3l + 4)2l³

Divisor = 2l²

=

= (2l² – 3l + 4)l

= (2l³ –2l² + 4l)

Hence dividing 4(l⁵ – 6l⁴ + 8l³) by 2l² gives out (2l³ –2l² + 4l)

In the given term

Dividend = 15 (a³b²c²– a²b³c² + a²b²c³)

Take out the common part in binomial term

= 3 × 5(a × a × a × b × b × c × c– a × a × b × b × b × c × c + a × a × b × b × c × c × c )

= 15 a²b²c²(a – b + c)

Divisor = 3abc

=

= 5abc[a-b + c]

= [5a²bc– 5ab²c + 5abc²]

Hence dividing 15 (a³b²c²– a²b³c² + a²b²c³) by 3abc gives out [5a²bc– 5ab²c + 5abc²]

In the given term

Dividend = (3p³– 9p²q - 6pq²)

Take out the common part in binomial term

= (3 × p × p × p– 3 × 3 × p × p × q - 2 × 3 × p × q × q )

= 3 × p(p²– 3pq - 2q²)

Divisor = (–3p)

=

= (-1) (p²– 3pq - 2q²)

= (2q² + 3pq - p²)

Hence dividing (3p³– 9p²q - 6pq²) by (–3p) gives out (2q² + 3pq - p²)

In the given term

Dividend = ( a²b²c² + ab²c²)

Take out the common part in binomial term

= ( × a × a × b × b × c × c + × 2 × a × b × b × c × c )

= × a × b × b × c × c (a + 2)

= ab _�²c²(a + 2)

Divisor = abc

=

=

= bc(a + 2)

Hence dividing ( a²b²c² + ab²c²) by abc gives out bc(a + 2)

In the given term

Dividend = (49x – 63)

Take out the common part in binomial term

= (7 × 7 × x - 7 × 9)

= 7(7 × x - 9)

= 7(7x - 9)

Divisor = 7

=

= (7x - 9)

Hence dividing (49x – 63) by 7 gives out (7x - 9)

In the given term

Dividend = 12x (8x – 20)

Take out the common part in binomial term

= 2 × 2 × 3 × x(2 × 2 × 2 × x - 2 × 2 × 5 )

= 2 × 2 × 2 × 2 × 3 × x(2 × x - 5 )

= 48x (2x - 5 )

Divisor = 4(2x – 5)

= 12x

Hence divides 12x (8x – 20) by 4(2x – 5) gives out 12x

In the given term

Dividend = 11a³b³(7c – 35) ÷ 3a²b²(c – 5)

Take out the common part in binomial term

= 11 × a × a × a × b × b × b (7 × c - 5 × 7 )

= 11 × a × a × a × b × b × b × 7 (c - 5 )

= 77a³b³(c - 5)

Divisor = 3a²b²(c – 5)

=

= ab

Hence dividing 11a³b³(7c – 35) by 3a²b²(c – 5) gives out ab

In the given term

Dividend = 54lmn (l + m) (m + n) (n + l)

Divisor = 81mn (l + m) (n + l)

=

= l(m + n)

Hence dividing 54lmn (l + m) (m + n) (n + l) by 81mn (l + m)(n + l) gives out l(m + n)

In the given term

Dividend = 36 (x + 4) (x² + 7x + 10)

Divisor = 9 (x + 4)

=

= 4(x² + 7x + 10)

= (4x² + 27x + 40)

Hence dividing 36 (x + 4) (x² + 7x + 10) by 9 (x + 4) gives out

(4x² + 27x + 40)

In the given term

Dividend = a (a + 1) (a + 2) (a + 3)

Divisor = a (a + 3)

=

= (a + 1) (a + 2)

Hence dividing a (a + 1) (a + 2) (a + 3) by a (a + 3) gives out

(a + 1) (a + 2)

In the given term

Dividend = (x² + 7x + 12)

The given expression looks as

x² + (a + b)x + ab

where a + b = 7; and ab = 12;

factors of 12 their sum

1 × 12 1 + 12 = 13

6 × 2 2 + 6 = 8

4 × 3 4 + 3 = 7

∴ the factors having sum 7 are 4 and 3

x² + 7x + 12 = x² + (4 + 3)x + 12

= x² + 4x + 3x + 12

= x(x + 4) + 3(x + 4)

= (x + 4)(x + 3)

Divisor = (x + 3)

=

= (x + 4)

Hence dividing (x² + 7x + 12) by (x + 3) gives out (x + 4)

In the given term

Dividend = (x² - 8x + 12)

The given expression looks as

x² + (a + b)x + ab

where a + b = -8; and ab = 12;

factors of 12 their sum

-1 × -12 -1-12 = -13

-6 × -2 -2-6 = -8

-4 × -3 -4-3 = -7

∴ the factors having sum 7 are 4 and 3

x² - 8x + 12 = x² + (-6-2)x + 12

= x² - 6x - 2x + 12

= x(x - 6) -2(x - 6)

= (x - 6)(x - 2)

Divisor = (x - 6)

=

= (x – 2)

Hence dividing (x² – 8x + 12) by (x – 6) gives out (x – 2)

In the given term

Dividend = (p² + 5p + 4)

The given expression looks as

x² + (a + b)x + ab

where a + b = 5; and ab = 4;

factors of 4 their sum

1 × 4 1 + 4 = 5

2 × 2 2 + 2 = 4

∴ the factors having sum 5 are 4 and 1

(p² + 5p + 4) = p² + (4 + 1)p + 4

= p² + 4p + p + 4

= p(p + 4) + 1(p + 4)

= (p + 1)(p + 4)

Divisor = (p + 1)

=

= (p + 4)

Hence dividing (p² + 5p + 4) by (p + 1) gives out (p + 4)

In the given term

Dividend = 15ab (a²–7a + 10)

The given expression (a²–7a + 10) looks as

x² + (a + b) x + ab

where a + b = -7; and ab = 10;

factors of 10 their sum

-1 × -10 -1-10 = -11

-2 × -5 -2-5 = -7

∴ the factors having sum -7 are -2 and -5

(a²–7a + 10) = a² + (-2-5)a + 10

= a²–5a – 2a + 10

= a(a – 5) – 2(a – 5)

= (a – 5)(a – 2)

Divisor = 3b (a – 2)

=

= 5a(a – 5)

Hence dividing 15ab (a²–7a + 10) by 3b (a – 2) gives out 5a(a – 5)

In the given term

Dividend = 15lm (2p²–2q²)

In given expression (2p²–2q²)

Take out the common factor in binomial term

⇒ (2 × p × p – 2 × q × q)

→ 2(p² – q²)

Both terms are perfect square

⇒ p² = p × p

⇒ q² = q × q

∴ (p² – q²) Seems to be in identity a²-b² = (a + b)(a-b)

Where a = p and b = q;

p² – q² = (p + q)(p – q)

Hence the factors of p² – q² are (p + q) and (p – q)

Divisor = 3l (p + q)

=

=

= 10m(p – q)

Hence dividing 15lm (2p²–2q²) by 3l (p + q) gives out 10m(p – q)

In the given term

Dividend = 26z³(32z²–18)

Take out the common factor in binomial term

⇒ 2 × 13 × z × z × z (2 × 2 × 2 × 2 × 2 × z × z – 2 × 3 × 3)

⇒ 2 × 2 × 13 × z × z × z (2 × 2 × 2 × 2 × z × z – 3 × 3)

⇒ 52z³(16z² – 9)

In given expression (16z² – 9)

Both terms are perfect square

⇒ 16z² = 4z × 4z

⇒ 9 = 3 × 3

∴ (16z² – 9) Seems to be in identity a²-b² = (a + b)(a-b)

Where a = 4z and b = 3;

(16z² – 9) = (4z + 3)(4z – 3)

Hence the factors of (16z² – 9)are (4z + 3) and (4z – 3)

Divisor = 13z²(4z – 3)

=

=

= 4z(4z + 3)

Hence dividing 26z³(32z²–18) by 13z²(4z – 3) gives out 4z(4z + 3)

If LHS is

3(x – 9)

Then RHS would be

⇒ 3(x – 9)

= 3 × x – 3 × 9

= 3x – 27

The error is 27 instead of 9

Hence 3(x – 9) = 3x – 27

If LHS is

x(3x + 2)

Then RHS would be

⇒ x(3x + 2)

= 3 × x × x – 2 × x

= 3x² – 2x

The error is 2x instead of 2

Hence x(3x + 2) = 3x² + 2x

If LHS is

2x + 3x

Then RHS would be

⇒ 2x + 3x

= x(2 + 3)

= 5x

The error is 5x instead of 5x²

Hence 2x + 3x = 5x

If LHS is

2x + x + 3x = 5x

Then RHS would be

⇒ 2x + x + 3x

= x(2 + 1 + 3)

= 6x

The error is 6x instead of 5x

Hence 2x + x + 3x = 6x

If LHS is

4p + 3p + 2p + p – 9p

Then RHS would be

⇒ 4p + 3p + 2p + p – 9p

= p(4 + 3 + 2 + 1–9)

= p(10–9)

= p

The error is p instead of 0

Hence 4p + 3p + 2p + p–9p = p

If RHS is

6xy

Then LHS would be

⇒ 6xy

= 2 × 3 × x × y

= 3 × x × 2 × y

= 3x × 2y

The error is sign of multiplication instead of sign of addition

Hence 3x × 2y = 6xy

If LHS is

(3x)² + 4x + 7

Then RHS would be

⇒ (3x)² + 4x + 7

= 3² × x² + 4x + 7

= 9x² + 4x + 7

The error is 9x² instead of 3x²

Hence (3x)² + 4x + 7 = 9x² + 4x + 7

If LHS is

(2x)² + 5x

Then RHS would be

⇒ (2x)² + 5x

= 2² × x² + 5x

= 4x² + 5x

The error is 4x² instead of 4x

Hence (2x)² + 5x = 4x² + 5x

If LHS is

(2a + 3)²

Then RHS would be

⇒ (2a + 3)²

= (2a)² + 3² + 2 × 2a × 3

= 4a² + 9 + 12a

= 4a² + 12a + 9

The error is 4a² instead of 2a² and 12a instead of 6a

Hence = (2a + 3)² = 4a² + 9 + 12a

If LHS is

x² + 7x + 12

Then RHS would be

⇒ x² + 7x + 12

Putting x = (-3)

= (–3)² + 7 (–3) + 12

= 9 + (-21) + 12

= 21-21

= 0

The error is (-21) instead of 4 and end result 0 instead of 25

Hence putting x = (-3) in x² + 7x + 12 results to 0

If LHS is

x²– 5x + 6

Then RHS would be

⇒ x²– 5x + 6

Putting x = (-3)

= (–3)² –5 (–3) + 6

= 9 + 15 + 6

= 30

The error is + 15 instead of (-15) and end results to 30 instead of 0

Hence putting x = (-3) in x²– 5x + 6 results to 30

If LHS is

x² + 5x

Then RHS would be

⇒ x² + 5x

Putting x = (-3)

= (–3)² + 5 (–3)

= 9 + (-15)

= -6

The error is ( + 9) instead of (-9) and end results to (-6) instead of (-24)

Hence putting x = (-3) in x² + 5x results to (-6)

If LHS is

(x – 4)²

Then RHS would be

⇒ (x – 4)²

= (x)² + 4² – 2 × x × 4

= x² + 16 – 8x

The error is x² + 16 – 8x instead of x² – 16

Hence (x – 4)² = x² + 13 – 8x

If LHS is

(x + 7)²

Then RHS would be

⇒ (x + 7)²

= (x)² + 7² + 2 × x × 7

= x² + 49 + 14

The error is x² + 14x + 49 instead of x² + 49

Hence (x + 7)² = x² + 14x + 49

For getting in the equation

(a² – b² ) = (a + b)(a-b)

RHS would be

3a² – 4b²

Then LHS would be

⇒ 3a² – 4b²

= (3a – 4b)(3a + 4b)

The error is (a – b) instead of (3a – 4b)

3a² – 4b² instead of 3a² – 4a²

Hence 3a² – 4b² = (3a – 4b)(3a + 4b)

If LHS is

(x + 4) (x + 2)

Then RHS would be

⇒ (x + 4) (x + 2)

= x² + 4 × x + 2 × x + 2 × 4

= x² + 4x + 2x + 8

= x² + 6x + 8

The error is x² + 6x + 8 instead of x² + 8

Hence (x + 4) (x + 2) = x² + 6x + 8

If LHS is

(x – 4) (x – 2)

Then RHS would be

⇒ (x – 4) (x – 2)

= x² – 4 × x – 2 × x + (-2) × (-4)

= x² – 4x – 2x + 8

= x² – 6x + 8

The error is x² – 6x + 8 instead of x² – 8

Hence (x – 4) (x – 2) = x² – 6x + 8

If LHS is

5x³ ÷ 5x³

Then RHS would be

⇒ 5x³ ÷ 5x³

=

= 1

The error is1 instead of 0

Hence 5x³ ÷ 5x³ = 1

If LHS is

(2x³ + 1) ÷ 2x³

Then RHS would be

⇒ (2x³ + 1) ÷ 2x³

=

=

=

The error is instead of 1

Hence (2x³ + 1) ÷ 2x³ =

If LHS is

(3x + 2) ÷ 3x

Then RHS would be

⇒ (3x + 2) ÷ 3x

=

=

=

The error is instead of

Hence (3x + 2 )÷ 3x =

If LHS is

For the complete and perfect division

There must be 3x instead of x

(3x + 5)÷3x

Then RHS would be

⇒ (3x + 5)÷3x

=

=

=

The error is instead of 5 and 3x instead of x

Hence = (3x + 5)÷3x =

If LHS is

Then RHS would be

⇒

= x +

= x + 1

The error is x + 1 instead of x + 1

Hence x + 1